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10 votes
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Projective tensor squares of uniform algebras

In discussion with a colleague recently (Jan 2017), $\newcommand{\AD}{A({\bf D})}\newcommand{\CT}{C({\bf T})}$ I was reminded that if $A(D)$ denotes the disc algebra and $\iota: \AD\to \CT$ is the ...
Yemon Choi's user avatar
  • 25.8k
6 votes
0 answers
117 views

Homomorphisms from BV

Denote by $\mathsf{BV}(\mathbb T)$ the Banach space of functions on the circle with bounded variation which is a Banach algebra under the pointwise product. Is there a surjective homomorphism from $\...
Maciej Ciechowski's user avatar
4 votes
0 answers
264 views

Is the Gelfand transform strictly continuous?

Let $M$ be the Banach algebra of measures on the circle with $L_1$ naturally sitting as a closed ideal of $M$. Then $M$ carries the strict topology implemented by the family of seminorms $\|\mu\|_f = \...
Jan_Ch.'s user avatar
  • 113
4 votes
0 answers
551 views

$f,g , |f|f, |g|g \in A(\mathbb R) \ \text{(Banach algebra)} \implies \left\|f|f|- g|g|\right\|\leq C \left \|f-g\right \|$?

Let $f\in L^{1}(\mathbb R)$ and it Fourier transform, $\hat{f} (y) : = \int _ {\mathbb R} f(x) e^{-2\pi i x\cdot y} dx ; y \in \mathbb R ;$ and consider Fourier algebra $$A(\mathbb R):= \{f\in L^{1}(...
Inquisitive's user avatar
  • 1,051
3 votes
0 answers
286 views

Tauberian theorem from generalized Gelfand transform

Wiener's theorem gives the necessary and sufficient conditions for the set of translates of a set of functions to be dense in $L^1(\mathbb{R}^n)$, which translates algebraically into a statement about ...
Adam Hughes's user avatar
  • 1,049
2 votes
0 answers
81 views

An square root of the multiplicative operator on $\ell^1(\mathbb{Z}_n)$

Let us consider the finite group algebra $\ell^1(\mathbb{Z}_n)$. Let $x=(x_0,\cdots,x_{n-1})$ in $\ell^1(\mathbb{Z}_n)$ and define $$M_x: \ell^1(\mathbb{Z}_n)\to \ell^1(\mathbb{Z}_n) : M_x(a)=a*x$$ ...
ABB's user avatar
  • 4,058
2 votes
0 answers
149 views

A closed ideal in $L^1(T)$

Let $\mathbb{T}$ be the unit circle and consider the convolution group algebra $L^1(\mathbb{T})$. Let $I_n$ be the closed ideal generated by the polynomial $p_n(z)=z^n-1$ in $L^1(\mathbb{T})$. Let $I=...
ABB's user avatar
  • 4,058
2 votes
0 answers
276 views

pointwise limit of uniformly bounded sequence in $A(\mathbb T)$ is again in $A(\mathbb T)$?

Let $\mathbb T$ be a circle group, and $\hat{f}(n)= \frac{1}{2\pi}\int_{0}^{2\pi} f(t) e^{-int} dt;$ $(n\in \mathbb Z, f\in L^{1} (\mathbb T)).$ Put $A(\mathbb T)= \{f\in C(\mathbb T): \hat{f}\in \ell^...
Inquisitive's user avatar
  • 1,051
1 vote
0 answers
102 views

How do functions operates in a Fourier algebra $A^{q}(\mathbb T)$?

We put , $A^{q}(\mathbb T)= \{ f\in L^{q}(\mathbb T): \hat{f}\in \ell^{q}(\mathbb Z) \}.$ By Helson-Kahane-Katznelson-Rudin Theorem, it follows that, "Let $F$ be a function on $\mathbb C$ and if $F(f)...
Inquisitive's user avatar
  • 1,051
1 vote
0 answers
269 views

$L^{p}(\mathbb R)\subset L^{1}(\mathbb R) \ast L^{p}(\mathbb R), (1< p< \infty)$?

Let $\mathbb T$ be a circle group. In 1939, Salem, has shown that, every member of $L^{1}(\mathbb T)$ can written as a product(convolution) some other two members of $L^{1}(\mathbb T),$ that is, $L^...
Inquisitive's user avatar
  • 1,051
1 vote
0 answers
187 views

Injective modules over Fourier algebra

Is there any article on injective modules over Fourier Algebras? Do we have anything about injectivity of $A(G)$ as a $A(G)$-bimodule?
Zora's user avatar
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